3 Answers
3

No, at least not with the usual calculus notion of integral (with Lebesgue measure in the background) and $f$ being bounded. Let $(x_n)$ be a nonnegative sequence converging to $x$ and $f$ be a bounded function such that the integral $I_x=\int_0^xf(y)~dy$ is finite and well defined for all $x\geq 0$.

Suppose $I_{x_n}$ does not converge to $I_x$. Then there is an $\epsilon>0$ such that for all $\delta>0$, there is $x'$ such that $|I_x-I_{x'}|>\epsilon$ and $|x-x'|<\delta$. Let $B>0$ be such that $-B< f(x)<B$ for all $x\geq 0$. Then the integral $$\bigg|\int_{x-\delta}^{x+\delta}f(y)~dx\bigg|\leq 2\delta B$$ for all $\delta>0$. In particular for, $\delta<\epsilon/(B2)$, the integral can not vary that much and we get a contradiction.

A continuous function on a closed and bounded interval is of course already bounded.

In fact boundedness is not necessary either; if $f$ is Lebesgue integrable on $[0,b]$ then we can use dominated convergence to show $\int_0^x f$ is continuous on $[0,b]$.
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Nate EldredgeAug 14 '12 at 13:03

Let $f$ be a function and denote $F(x)=\int_0^xf(t)\,dt$. For $F$ to be well defined, $f$ must be integrable (e.g. in the Riemann sense). It is a fact that in this case $F$ is always continuous. Here is a proof:

Since $f$ is integrable, it is bounded: $m\leq f(x)\leq M$ for some $m,M$ and all $x$ in a suitable region. Then, taking $\varepsilon>0$ and assuming $|x-y|<\frac{\varepsilon}{\max\{|m|,|M|\}}$, we have
$$|F(x)-F(y)|=\left|\int_0^xf(t)\,dt-\int_0^yf(t)\,dt\right|=\left|\int_y^xf(t)\,dt\right|
\leq \max\{|m|,|M|\}|x-y|\leq \varepsilon $$
This shows that $F$ is continuous.